 Hello friends, myself Prashant Vishwanath Dinshati, assistant professor, department of civil engineering from Vajjan Institute of Technology, SolarPort. Today I am here to explain you about slope and deflection of cantilever beam with uniformly distributed load by double integration method. So, today's outcome at the end of this session students will be able to understand and find the values of slope and deflection of cantilever beams with uniformly distributed load for the following conditions that is when UDL is acting over whole span, when UDL is acting for a distance a from a fixed end. Now, before we start we will see what is a beam? A beam is a structural element that primary resist load applied laterally to the beams axis means this load is applied laterally and its mode of deflection is primarily by bending action. The beams may be classified depending upon support conditions that is cantilever beam, simply supported beam, overhanging beam, fixed beam and continuous beam. So, what is slope and deflection and why it is necessary? So, design of every beam is frequently governed by rigidity rather than its strength. So, every building code specifies limits on deflection as well as stresses. So, excessive deflection nor beam of a beam not only in visually disturbing but also may cause damage to other parts of buildings that is why the study of slope and deflection is very necessary. So, we will see what is deflection of beam? The deflection at any point on the axis of beam is the distance between its position before and after loading. So, when the beam is initially it is a straight and after loading it gets deflected to this shape and the distance between this deflected initially and after loading this is the deflection of beam and it is here greater at free end and smaller at the fixed end and 0 at fixed end. Slope of the beam, the slope of at any section in a deflected beam is defined as angle in radiance which the tangent at the section makes with original axis of the beam. So, now this is the deflected shape and here I have drawn a tangent and this theta is the slope of a beam. Cantilever beam is fixed at one end and free at the other end as shown in this figure. So, there are various method of finding slope and deflection for cantilevers, they are double integration method, maculis method and moment area method. So, we are dealing with double integration method. So, how the equation of double integration method came? So, if we see the deflected shape of the beam it is having a curvature and it is having at some point origin and that is having a radius. So, the curvature of the deflected beam is given by the flexural equation that is m upon i is equals to e by r. Therefore, m upon e i is equals to 1 by r. So, this is equation A and for practically beam, so this 1 upon r is denoted by d 2 y by dx square. So, this is equation B. So, equating equation A and B we get m upon e i is equals to d 2 y by dx square. Therefore, m is equals to e i d 2 y by dx square that is equation 1. So, now you will see the slope and deflection of cantilever with uniformly distributed load by double integration method. So, here is a beam which is having a u d l over a whole span and this is the deflected shape of the beam and here it is the y b deflection and it is the slope here theta b. So, that we are going to find. So, now for finding slope and deflection we will consider a section at a distance x from fixed end and the remaining portion is l minus x as the total length of the beam is l and the deflection here it is y b. So, now we will take a bending moment at this section. So, it is equals to this load into this distance. So, the u d l moment at section x is equals to minus w into this l minus x and it is acting at a distance of l minus x by 2. So, this is equation number 2 this minus sign I have given due to hogging bending moment, but we know from equation 1 m upon e i is m is equals to e i d 2 y by dx square now putting the value of m. So, we get e i d 2 y by dx square is equals to minus w into bracket l minus x into l minus x by 2. So, this we will get minus w by 2 into bracket l minus x square. So, integrating this above equation we get e i d y by dx is equals to minus w by 2 into l minus x cube by 3 into minus 1 plus c 1 that is equation number 3 this plus c 1 is the constant of integration. So, integrating again this equation number 3 we get e y y is equals to w by 6 into l minus x raise to 4 by 4 into minus 1 plus c 1 into x plus c 2. So, c 2 is again the constant for integration. So, this is the equation number 4 and c 1 and c 2 are the constant of integration which can be obtained from boundary conditions. So, what are boundary conditions for a cantilever beam? Here you pause the video and try to answer the boundary conditions try to write it on a paper then continue. Now, as the support is fixed the values obtained from boundary condition are at x is equals to 0 deflection y is 0 at x is equals to 0 dy by dx slope is 0. So, now substituting this x is equals to 0 and y is equals to 0 in equation number 4 we get 0 is equals to minus w by 24 into l minus 0 raise to 4 plus c 1 into 0 plus c 2. So, from this we get c 2 is equals to w l 4 by 24. Again by substituting at x is equals to 0 dy by dx is also 0 in equation number 3 we get 0 is equals to w by 6 into l minus 0 raise to 3 plus c 1 therefore, c 1 is minus w l cube by 6. Now, substituting the value of c 1 is equals to minus w l cube by 6 in equation 3 we get e i dy by dx is equals to w by 6 into bracket l minus x cube minus w l cube by 6. This is equation number 5 this equation is known as slope equation. So, from this equation we can find the slope at any point on the cantilever by just substituting the value of x the slope is maximum at free end let it be theta b now at free end x is equals to l. So, putting the value of x is equals to l in this above equation number 5 we get e i theta b is equals to w by 6 into bracket l minus l raise to 3 minus w l cube by 6. So, this I have replaced here x by l. So, this after solving we get it is equals to minus w l cube by 6 therefore, theta b is equals to minus w l cube by 6 e i. This is equation number 5 a now here the negative sign shows that the angle is in anticlockwise direction. Now, again substituting the value of c 1 is equals to minus w l cube by 6 and c 2 is equals to w l 4 by 24 in equation number 4 we get e i y is equals to minus w l minus w by 24 into bracket l minus x raise to 4 minus w l cube by 6 into x plus w l 4 by 24. So, this is equation number 6. So, now this equation is known as deflection equation. So, here also putting the value of x we can find the deflection at any point on a beam. So, let us find the deflection at free end that is y b. So, deflection will be maximum at free end. So, putting the value of x is equals to l in this above equation number 6 we get e i y b is equals to minus w by 24 into bracket l minus l raise to 4 minus w l cube by 6 into l plus w l 4 by 24. So, now here I have replaced this x with the value of l. So, solving this equation we get y b is equals to minus w l 4 upon 8 e i this is equation number 6 a. So, this deflection is again having a negative sign as the deflection is downwards. Now, we will see the slope and deflection for the cantilever with uniformly distributed load at a distance a from the fixed end. So, now this load is up to only some distance and the next distance is having no load. So, this distance is a and this remaining distance is l minus a. So, now let the theta c is equals to slope at this point c d y by d x is at c y c is the deflection at c and y b is the deflection at b. So, now the portion a c of the cantilever may be taken similar to the cantilever which we have explained with the UDL having a full load over the whole span. So, if we neglect this second part. So, we are having as the case we studied that is load is acting over the whole span instead of l we are replacing l by a. So, we get theta c d y by d x at c is equals to w a cube by 6 e i and y c is equals to w a 4 divided by 8 e i. Now the beam will bend only between a to c portion and the remaining portion it will remain straight as no bending moment is there in the in this portion. So, theta c is equals to theta b is equals to w a cube by the 6 e i and y b is equals to y c plus theta c into bracket l minus a. So, this putting theta c value we get y b is equals to w a 4 divided by 8 e i plus w a cube divided by 6 e i into bracket l minus a because theta c we have already calculated previously w a cube divided by 6 e i. So, these are the references which I have referred. Thank you.